LeetCode: Count and Say [037]

【题目】

The count-and-say sequence is the sequence of integers beginning as follows:

1, 11, 21, 1211, 111221, ...

1

is read off as

"one
1"

or

11

.

11

is read off as

"two
1s"

or

21

.

21

is read off as

"one
2

, then

one 1"

or

1211

.

Given an integer n, generate the nth sequence.

Note: The sequence of integers will be represented as a string.

【题意】

有一个这样的序列，序列中的每个数都是从前一个数通过count-and-say规则生成。

所谓count-and-say就是数数然后读出来，按照读法来组织新的数。举个例子来说

1 读成“1个1” ==> 11

11 读成“2个1” ==> 21

21 读成“1个2，1个1” ==> 1211

目标是返回序列中的第n个数

【思路】

按规则顺序生成序列中的前n个数

【代码】

class Solution {
public:
string getNext(string integer){
int count=0;
char digit='\0';
string newInteger="";
for(int i=0; i<integer.length(); i++){
if(integer[i]!=digit){
if(digit!='\0'){
char countChar='0'+count;
newInteger+=countChar;
newInteger+=digit;
}
count=1;
digit=integer[i];
}
else{
count++;
}
}
if(digit!='\0'){
char countChar='0'+count;
newInteger+=countChar;
newInteger+=digit;
}
return newInteger;
}
string countAndSay(int n) {
int count=1;
string integer="1";
if(n<1)return "";   //考虑一些非法值，题目没有规定，那就设定为空
if(n==1)return integer;

while(count<n){
integer=getNext(integer);
count++;
}
return integer;
}
};